- $(\Delta T)_m = Tm_1 - Tm_2 = 110^\circ C = 110k$
- Heat lost by the metal block
- $Q_m = C_m (0.2kg)(110k)= 22C_m J \quad ... (1)$
- $ \Delta E (\Delta T)_{W,C} = 40^\circ C - 27^\circ C = 13^\circ C = 13K$
- $Q_T = Q_W + Q_C = M_WC_W(\Delta T) + M_CC_C(\Delta T)$
- $(13.585 + 0.70^3) \times 10^3 J \quad ... (2)$
- $Q_T = Q _m$
- $C_m = 0.649 \times 10^3(J kg^{-1}k^{-1})$
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-11-unit-11-chapter-02-thermal-properties-of-matter-2-lecture-2_4-qdeyud3ck-0-8.jpg)